Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
template <typename A, typename B>
inline char smax(A &a, const B &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename A, typename B>
inline char smin(A &a, const B &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename I>
inline void read(I &x) {
  int f = 0, c;
  while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
  x = c & 15;
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
  f ? x = -x : 0;
}
const int N = 1000 + 7;
int n;
inline void work() {
  printf("%d\n", n / 2 + 1);
  int m = n / 2 + 1;
  for (int i = 1; i <= n / 2; ++i)
    if (i & 1)
      printf("%d %d\n", i, 1);
    else
      printf("%d %d\n", 1, i);
  for (int i = n / 2 + 1; i <= n; ++i)
    if (i & 1)
      printf("%d %d\n", m, i - m + 1);
    else
      printf("%d %d\n", i - m + 1, m);
}
inline void init() { read(n); }
int main() {
  init();
  work();
  fclose(stdin), fclose(stdout);
}